The right most square is cartesian in all categories since \(x \in U\)  is contained in the isomorphism locus, and the composite of the two left most squares are cartesian by definition. The rest of the squares are cartesian by standard arguments. Furthermore, letting \(C^{\prime }\rightarrow \mathcal {M}(\tilde{X},\beta ^{\prime })\)  and \(C \rightarrow \mathcal {M}(X,\beta )\)  be the universal curves, we know by step 2 that the stabilization morphism \(C^{\prime } \rightarrow C\)  is an isomorphism locally around the section \(x_{out}: \mathcal {M}(\tilde{X},\beta ^{\prime }) \rightarrow C^{\prime }\) . Hence, the divisor \(\mathcal {O}_C(-x_{out})\)  pulls back to \(\mathcal {O}_{C^{\prime }}(-x_{out})\) . This fact, together with the main result of [1]}, says that the obstruction theory for \(\mathcal {M}(\tilde{X},\beta ) \rightarrow \mathfrak {M}^{ev}(\tilde{\mathcal {X}},\beta )\)  is the pullback along the stabilization morphism of the obstruction theory for \(\mathcal {M}(X,\beta ) \rightarrow \mathfrak {M}^{ev}(\mathcal {X},\beta )\) . By definition of the perfect obstruction theories on \(\mathcal {M}(X,\beta )_x\)  and \(\mathcal {M}(\tilde{X},\beta ^{\prime })_x\)  and the commutativity of the left most square, the obstruction theory on \(\mathcal {M}(\tilde{X},\beta ^{\prime })_x\)  pulls back from the obstruction theory defined on \(\mathcal {M}(X,\beta )_x\) . In order to conclude \(\pi _*([\mathcal {M}(\tilde{X},\beta ^{\prime })_x]^{vir}) = [\mathcal {M}(X,\beta )_x]^{vir}\)  from this fact, we need to compute the virtual degree of \(\mathcal {M}(\tilde{X},\beta )_x \rightarrow \mathcal {M}(X,\beta )_x\) .
